Lets S={z in C:|z-1|=1 and (sqrt{2}-1)(z+bar{ | vedclass

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Question

Let $Z=x+i y$

Then $(x-1)^2+y^2=1 \rightarrow(1)$

$(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$

$\quad \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \righta

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Let $Z=x+i y$

Then $(x-1)^2+y^2=1 \rightarrow(1)$

$(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$

$\quad \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)$

Solving $(1) and (2)$ we get

Either $\mathrm{x}=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow$ $(3)$

On solving $(3)$ with $(2)$ we get

For $x=1 \Rightarrow y=1 \Rightarrow Z_2=1+i$  and  for

$x=\frac{1}{2-\sqrt{2}} \Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}$

Now

$\left|\sqrt{2} z_1-z_2\right|^2$

$=\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2$

$=(\sqrt{2})^2$

$=2$

Lets $S=\{z \in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}$. Let $\mathrm{z}_1, \mathrm{z}_2$ $\in S$ be such that $\left|z_1\right|=\max _{z \in S}|z|$ and $\left|z_2\right|=\min _{z \in S}|z|$. Then $\left|\sqrt{2} z_1-z_2\right|^2$ equals :

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